
A proton (mass \[ = 1.67 \times {10^{ - 27}}kg\;\]and charge \[ = 1.6 \times {10^{ - 19}}C\]) enters perpendicular to a magnetic field of intensity \[2\;weber\;/{m^2}\;\]with a velocity \[3.4 \times {10^7}\;m/sec\]. Find the acceleration of the proton .
A. \[6.5 \times {10^{15}}m/{s^2}\;\]
B. \[6.5 \times {10^{13}}m/{s^2}\;\]
C. \[6.5 \times {10^{11}}m/{s^2}\;\]
D. \[6.5 \times {10^9}m/{s^2}\;\]
Answer
162.6k+ views
Hint: In the given question, we need to find the acceleration of a proton. For this, we need to use the formula for force experienced by a charged particle in an external magnetic field to get the desired result.
Formula used:
The following formula is used for solving the given question.
The force experienced by the proton due to the magnetic field is given by
\[F = m \times a = q \times v \times B\]
Here, \[F\] is the force, \[m\] is the mass, \[a\] is the acceleration, \[q\] is the charge, \[v\] is the velocity and \[B\]is the magnetic field.
Complete answer:
We know that the force experienced by the proton as a result of the magnetic field causes it to accelerate.
Mathematically, it is given by
\[F = m \times a = q \times v \times B\]
Here, \[F\] is the force, \[m\] is the mass, \[a\] is the acceleration, \[q\] is the charge, \[v\] is the velocity and \[B\] is the magnetic field.
So, we get
\[a = \dfrac{{q \times v \times B}}{m}\]
But \[q = 1.6 \times {10^{ - 19}}C\], \[v = 3.4 \times {10^7}\;m/sec\], \[B = 2\;weber\;/{m^2}\;\]and \[m = 1.67 \times {10^{ - 27}}kg\;\]
So, we get
\[a = \dfrac{{1.6 \times {{10}^{ - 19}} \times 3.4 \times {{10}^7} \times 2}}{{1.67 \times {{10}^{ - 27}}}}\]
By simplifying, we get
\[a = \dfrac{{1.6 \times {{10}^{ - 19}} \times 3.4 \times {{10}^7} \times 2}}{{1.67 \times {{10}^{ - 27}}}}\]
This gives \[a = 6.5 \times {10^{15}}m/{s^2}\;\]
Hence, the acceleration of the proton is \[a = 6.5 \times {10^{15}}m/{s^2}\;\].
Therefore, the correct option is (A).
Note: Moving charges experience forces from magnetic fields. This is among the most fundamental forces known. The magnetic force on a moving charge is directed perpendicular to the plane produced by velocity and magnetic field and fulfills the right-hand hand rule. Many students make mistakes in the calculation part as well as writing simplifying the power of ten. This is the only way, through which we can solve the example in the simplest way. Also, it is essential to do calculations carefully to get the correct value of the acceleration of a proton.
Formula used:
The following formula is used for solving the given question.
The force experienced by the proton due to the magnetic field is given by
\[F = m \times a = q \times v \times B\]
Here, \[F\] is the force, \[m\] is the mass, \[a\] is the acceleration, \[q\] is the charge, \[v\] is the velocity and \[B\]is the magnetic field.
Complete answer:
We know that the force experienced by the proton as a result of the magnetic field causes it to accelerate.
Mathematically, it is given by
\[F = m \times a = q \times v \times B\]
Here, \[F\] is the force, \[m\] is the mass, \[a\] is the acceleration, \[q\] is the charge, \[v\] is the velocity and \[B\] is the magnetic field.
So, we get
\[a = \dfrac{{q \times v \times B}}{m}\]
But \[q = 1.6 \times {10^{ - 19}}C\], \[v = 3.4 \times {10^7}\;m/sec\], \[B = 2\;weber\;/{m^2}\;\]and \[m = 1.67 \times {10^{ - 27}}kg\;\]
So, we get
\[a = \dfrac{{1.6 \times {{10}^{ - 19}} \times 3.4 \times {{10}^7} \times 2}}{{1.67 \times {{10}^{ - 27}}}}\]
By simplifying, we get
\[a = \dfrac{{1.6 \times {{10}^{ - 19}} \times 3.4 \times {{10}^7} \times 2}}{{1.67 \times {{10}^{ - 27}}}}\]
This gives \[a = 6.5 \times {10^{15}}m/{s^2}\;\]
Hence, the acceleration of the proton is \[a = 6.5 \times {10^{15}}m/{s^2}\;\].
Therefore, the correct option is (A).
Note: Moving charges experience forces from magnetic fields. This is among the most fundamental forces known. The magnetic force on a moving charge is directed perpendicular to the plane produced by velocity and magnetic field and fulfills the right-hand hand rule. Many students make mistakes in the calculation part as well as writing simplifying the power of ten. This is the only way, through which we can solve the example in the simplest way. Also, it is essential to do calculations carefully to get the correct value of the acceleration of a proton.
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